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Question: If in a voltaic cell 5 gm of zinc is consumed, then we get how many ampere hours ? (Given that E.C.E...

If in a voltaic cell 5 gm of zinc is consumed, then we get how many ampere hours ? (Given that E.C.E. of Zn is 3.387×10−73.387 \times 10^{- 7} kg/coulomb)

A

2.05

B

8.2

C

4.1

D

5×3.387×10−75 \times 3.387 \times 10^{- 7}

Answer

4.1

Explanation

Solution

m=Zit=Zqm = Zit = Zq; q=5×10−33.387×10−7amp−sec⁡q = \frac{5 \times 10^{- 3}}{3.387 \times 10^{- 7}}amp - \sec

or q=5×10−33.387×10−7×3600amp−hr=4.1q = \frac{5 \times 10^{- 3}}{3.387 \times 10^{- 7} \times 3600}amp - hr = 4.1